The projection of $ \vec{v_1} $ on the vector $ \vec{v_2} $ is given by

$$ \text{Proj}_{\vec{b}}{\vec{a}} = \dfrac{ \vec{a} \cdot \vec{b} }{ \| \vec{a} \|^2 } \vec{b}$$

First we find the dot product and magnitude of vector $ \, \vec{b} $:

$$ \begin{aligned}\vec{a} \cdot \vec{b} &= 5 \\[1 em]\| \vec{b} \| &= 5 \sqrt{ 2 } \\[1 em]\end{aligned} $$

Now we can find the projection

$$ \text{Proj}_{\vec{b}}{\vec{a}} = \dfrac{ 5 }{ \left( 5 \sqrt{ 2 } \right)^2 } \cdot \vec{b} = \dfrac{ 5 }{ 50 } \cdot \vec{b} = \frac{ 1 }{ 10 } \cdot \left(5,~-5\right) = \left(\dfrac{ 1 }{ 2 },~-\dfrac{ 1 }{ 2 }\right) $$

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